# Standard Deviation

If a question asked to find on a specific confidence level then we have that:

$P(-a \le \hat {p} - p \le a) = P(-a \le \frac{X_1 + ... +X_n - np}{n} \le a) = P(\frac {-a\sqrt n}{\sqrt {p(1-p)}} \le \frac {X_1 + ... +X_n - np}{\sqrt n \sqrt {p(1-p)}} \le \frac{a\sqrt n}{\sqrt {p(1-p)}}$

How were they able to find the standard deviation of $X_1$ to be $\sqrt {p(1-p)}$? I have not seen that formula before. I have only seen the case when

$$P(\frac{-a\sqrt n}{\sigma} \le \hat {p} - p \le \frac{a\sqrt n}{\sigma})$$

In what cases do we want to use $\sqrt {p(1-p)}$ from $\frac{-a\sqrt n}{\sigma}$ to find the standard deviation?

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The $X_i$ are presumably independent Bernoulli random variables with parameter $p$. Work out the mean and standard deviation of such a random variable (Hint: you should get $p$ and $\sqrt{p(1-p)}$. If you don't, try, try again. – Dilip Sarwate Oct 31 '12 at 2:49

In this answer it is derived that the expected value of $n$ Bernoulli trials is $n\color{#C00000}{p}$ and the variance is $n\color{#C00000}{p(1-p)}$. Thus, the standard deviation of a single trial is $\sqrt{p(1-p)}$.